A 41.12 g sample of a substance is initially at 29.5 °C. After absorbing 1075 J...
Homework Answers
heat absorbed = 1075 J
mass of sample = 41.12 g
temperature rise = 166.5 - 29.5 = 137 oC
Q = m Cp dT
1075 = 41.12 x Cp x 137
Cp = 0.191 J / g oC
specific heat = 0.191 J / g oC
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